Subject: [mg128504] Re: How accurate is the solution for high degree algebraic equation?

From: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>

Date: Thu, 25 Oct 2012 23:33:06 -0400 (EDT)

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I wanted to know all the solutions of f = (-z - 1)^d - (-z^d - 1)==0, where d=54.
I did the following:
d = 54; f = (-z - 1)^d - (-z^d - 1);
sol = NSolve[f == 0,z];
a = z /. sol;
So a is a set of solutions.
If I compute
f /. z -> a[[50]] // N
It returns a number very close to zero. This is natural.
But if I compute
f /. (z -> a[[1]]) // N
Then
Mathematica returns
12.0047 + 14.7528 I
I cannot say a[[1]] is a solution of f=0.
Many other elements in the solution set a does not seem to satisfy the equation.
Only the last few terms in a are satisfactory enough as solutions.
Is the degree too high?
Hi, Alexandra,
It seems that your equation has no solution in real numbers, if you are looking for real solutions, of course. Let me give the arguments.
Here is your function:
d = 54;
f[z_] := (-z - 1)^d - (-z^d - 1);
One can look, how it behaves close to zero. Evaluate this:
Plot[f[z], {z, -0.05, 0.1}]
It is easy to see that at least at 0<z<0.1 it is positive. Let us look, if it is monotone. Its derivative is:
D[f[z], z]
-54 (-1 - z)^53 + 54 z^53
Let us find the points, where this derivative might change its sign. This is defined by the equation
-54 (-1 - z)^53 + 54 z^53=0
And it is elementary to be solved by hand in real numbers:
z=-1/2
All other 52 solutions are complex and, thus, they are of no interest.
Since we have seen at the plot that f(z) increases at z>0, we must conclude that, at z=-1/2 the function f(z) changes its behaviour from decreasing to increasing. Easy to find
f[-1/2] // N
1.
That means that the function f(z) is found in the upper semi-plane and never reaches zero, its minimum value, fmin=1, being reached at z=-1/2.
Of course, the things change, if you are looking for complex solutions.
Have fun, Alexei
Alexei BOULBITCH, Dr., habil.
IEE S.A.
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e-mail: alexei.boulbitch at iee.lu